text:
How do you calculate the ground distance in a climb?
The 172M POH doesn't have a "Distance to climb" chart like the newer 172s. I just have the climb rate (Call it 600 ft/min) and so I estimated a time of 8.5 minutes. The problem is, even if I calculated the distance (about 1 NM), it's the vertical distance which I know, not the ground distance which I'm looking for. Do I have to use trigonometry to calculate my ground distance?
See page 65

You don't need to use trig if you have time and speed...  Think of it this way, if you were NOT  climbing, how much distance would you cover in 8.5 minutes at __ airspeed?

How precise do you need?

Actually, distance to climb involves 2 critical components of preflight planning, weight and weather.  Climb rate and distance to climb are heavily influenced by these factors.  "Simplifying" things does not equal skipping important steps, nor does it mean taking a little time to understand why a little "shorthand" works for a 172.  Distance to climb becomes important if terrain or something like a radio tower (windmills these days) is in the way.

One key point that has not been explicitly mentioned so far, though it's apparent from the example given in another answer, is that for light piston-engine airplanes, the angle of climb is so shallow that there's no need to go through the step of using trigonometry to calculate your climb angle to convert the speed and distance travelled along the flight path to the speed and distance travelled over the ground.  For example if you are climbing at 70 knots true airspeed, just assume that the horizontal component of the airspeed vector is also 70 knots.  Then you are ready to factor in the wind to calculate your groundspeed, and your distance travelled over the ground.
Here are some numbers to help make this point clearer-- in still air, to make the horizontal component of your airspeed vector be 5% smaller than your total airspeed vector, you'd need a climb angle of 18.2 degrees.  At 70 knots airspeed, this would require a climb rate of 22.9 knots, or 2210 feet per minute-- far beyond the capacity of a Cessna 172.

Your opening sentence mirrors my comment.  No need for trig on a light single, keep things simple.  You got my upvote.

(re "capacity" -- "capability" might be a better word)

You would typically use four pieces of information to calculate the distance to climb:

start altitude
altitude to reach
vertical speed
ground speed**

First you calculate the total altitude to gain:

altitude to gain = altitude to reach - start altitude

From the altitude to gain and the vertical speed, you can calculate the time to reach the target altitude

time to climb = altitude to gain / vertical speed

Now that we know the time it will take to climb, we can calculate the horizontal distance we will travel during the climb:

horizontal distance = time to climb * ground speed.

**

ground speed = true air speed*cos(climb angle)  +- along track wind;

Note that the cosine of a small angle is approximately 1. Since the climb angle is shallow, we can ignore the effect of the climb on the forward speed component. For example if the climb angle is 10 degrees, the forward component of the speed is cos(10°) = 0.985 ≈ 1. We can therefore leave out the trigonometry and work with the simplified:

ground speed = true air speed  +- along track wind;

Example:

start altitude = 1000 ft
altitude to reach = 4000 ft
vertical speed = 600 fpm
ground speed = 120 kts

Using the formulas above:

altitude to gain = 4000 - 1000 = 3000 ft
time to climb = 3000 / 600 = 5 minutes
horizontal distance = 120 * 5/60 = 10 NM

Note that this only works for relatively short climbs, otherwise effects of wind and air density will change both the ground speed and the climb performance during the climb. To estimate the climb distance for longer climbs, you can cut it into multiple segments of shorter climbs, and use appropriate vertical speeds and ground speed for each segments. However, this is not an quick exercise that is easily done in the cockpit.

Other answer have mentioned the key point that ground speed ~= true air speed +- wind, because climb angles are so shallow.  But your answer seems to omit it.  I'll suggest an edit to make this answer self-contained.

@PeterCordes, thank you for the suggested edit.

Well, you have your altimeter and your watch, so rate of climb can be exactly determined.
Distance covered on the ground is the $cosine$ of distance in the air.
Here you're in luck as the 172 has an angle of climb of around 6 degrees.
This is calculated as arc sin of 600 fpm climb/6000 feet per minute airspeed.
The cosine of 6 degrees is 99.5%, which means your ground distance is 99.5% of what you flew in the air.
The slightly trickier part is that true airspeed, not indicated, must be used for more precise measurements.
In a 172, it might be easier to look out the window for a landmark or use GPS for distance traveled.  This may be advisable considering head or tail winds will also skew any distance calculations.

Oh right, even though 1 part in 10 (climb speed as a fraction of airspeed) isn't tiny, 0.1^2 + x^2 = 1.0 is still only a tiny impact on the x component (for coordinates on a unit circle, which is why cos^2 + sin^2 = 1.0).  So yeah, cos(climb angle) = 0.99499 = sqrt(1 - 0.1^2)

While it is good to understand all the nuances involved in these sorts of questions, rule #1 when flying an airplane is to keep things simple so that you can maintain situational awareness of the big picture.  Approximations and rules of thumb can, and should be used liberally.
That said, at the climb angles of a typical light single GA airplane, small angle approximation lets us dispense with Trig and turn this into a simple rate problem:  If you are flying at 60 Knots for the 8.5 minutes in your example you will have covered 8.5 nautical miles.  Double the speed to 120 and you cover twice that distance, or 17 nautical miles. (don't forget to add/subtract wind for distance over the ground!)
Is that too fast or slow for your 172?  No problem; like Goldilocks pick an airspeed that’s just right such as 90 knots and split the difference for 12.75.  (Then round up to 13.  Or, easier mental math might be to round 8.5 down to 8, 1.5 of which is 12, then add a mile for some cushion to cover the 30 seconds you chopped off...)
SUMMARY:  You should be able to rapidly calculate in your head approximately how much distance you will cover in any given span of time.  (whether in climb, cruise, or descent)  For typical small general aviation aircraft the easy benchmarks to memorize are –

60 = 1 mile per minute
90 = 1.5 miles per minute
120 = 2 miles per minute

Knowing the above keeps the mental math simple, and allows you to approximate values in between.
It isn’t just useful for climbs, in fact estimating your Top of Descent (TOD) is just as important.  For example, let’s say you are cruising at 120 knots at 4000’, and planning a 500 FPM rate of descend to an airport at sea level.  It will take you 8 minutes to complete the descent, so you would want to start down around 16 miles out.
While trigonometry is very useful, I would strongly recommend against performing complex calculations in flight unless you are very comfortable with it.  And keep in mind that the combined effects of varying winds, TAS vs IAS, plus your ability to hold airspeed within +/- 5 KIAS will likely have a greater effect on actual vs calculated distance than the difference between the adjacent leg and the hypotenuse.


text:
Can three pilots accumulate total flight time while training?
Situation: On an aircraft with a flight deck large enough to fit multiple people, three pilots are onboard. 1 PIC in the left seat, and two certified copilots taking turns flying in the right seat for proficiency. When not in the right seat, the other pilot is seated nearby actively participating in training discussions and talking through training scenarios.
Question: Can the two copilots log the entire time of the flight towards their total flight time even when not seated in one of the two pilot seats? Of course they could only log SIC time when actually seated in the right pilot seat.

Which FAR are you reading that leads you to believe that might be possible?

61.1 states: Pilot time means that time in which a person -
(i) Serves as a required pilot flight crewmember. (ii) Receives training from an authorized instructor in an aircraft.
Re: (i) All three pilots are serving as required pilots. I understand only two pilots are required to fly this plane at minimum, but if the mission of the flight is training and proficiency, then are not the two trainee pilots required?
Re (ii): both pilots are receiving training at the same time by participating in discussions.

Also, I’ve read several comments on this forum about third pilots (whether two first officers on a flight or a first officer and a second officer) on the flight deck of long haul flights being able to log total time, but just not SIC time if not actually sitting in one of the two pilots seats; but those comments didn’t cite regulation or justification, so I’m just looking for more info regarding logging flight time in three pilot situations.

It doesn’t work that way.  That third guy is not required.

In the military we had a category called "special crew time" for any crewmember who wasn't a rated pilot, or for rated pilots not acting as PIC or SIC.  I don't know if it would be an issue with the FAA or a prospective employer if your log book PIC and SIC added up to less than total time, but I wouldn't even try.  (I certainly didn't count any of my mil crew time towards pilot ratings, nor did I put it on my resume...)

@MichaelHall my situation is actually about a military C130 flying with multiple junior pilots on board to take turns flying and observing. I didn’t think to ask my Question explicitly in terms of the FAA recognizing ‘special crew time,’ which is essentially what I’m describing.

@KRichter, well if you are active or reserve you probably should be asking your Ops Officer these sort of questions.  And I have no idea how the FAA treats special crew time, but I would presume it doesn't mean anything so I've never tried.  I lied above, on my flying resume I listed my crew time at the very end under "miscellaneous other work experience."  But then it wasn't mixed in with pilot time on the same tour.   FWIW...

@MichaelHall I’m researching for my Ops officer. It’s only recently we started flying with three pilots to spread out the training opportunities, and we’re just trying to figure out how to log this flight time appropriately per FAA rules.

Interesting, we used to fly with extra pilots all the time.  See my answer and let me know what you think.  What branch of service?

Your question of whether or not a copilot can log SIC time while not seated in a crew seat is an absolute NO.  They cannot log SIC time.

§61.51(f) Logging second-in-command flight time.
A person may log second-in-command time only for that flight time during which that person:

(1) Is qualified in accordance with the second-in-command requirements of § 61.55, and occupies a crewmember station in an aircraft that requires more than one pilot by the aircraft's type certificate;

(2) Holds the appropriate category, class, and instrument rating (if an instrument rating is required for the flight) for the aircraft being flown, and more than one pilot is required under the type certification of the aircraft or the regulations under which the flight is being conducted; or

(3) Serves as second in command in operations conducted in accordance with § 135.99(c) of this chapter when a second pilot is not required under the type certification of the aircraft or the regulations under which the flight is being conducted, provided the requirements in §61.159(c) are satisfied.

To log total flight time you must be a required crewmember.  A required crewmember is defined by the aircraft certification or by the operating rules.
There are no rules I am aware of that allows you to do this.
See 61.55(a)

A person may serve as a second-in-command of an aircraft type certificated for more than one required pilot flight crewmember or in operations requiring a second-in-command pilot flight crewmember only if that person holds:

thanks for post. My question isn’t whether the pilot not in a pilot seat can log SIC time; but whether the pilot not in a pilot seat can accumulate general pilot time or time towards their total flight time.
Pilots not in a pilot seat of long haul flights can still log total flight time. I’m wondering if the same could be done in a multiple pilot training mission/situation. Thanks again.

@757toga the second part: I’m asking how the ‘total flight time’ would apply and be logged per FARs. I am beginning to see that only nonflying pilots on long-haul flights are allowed to log total pilot time when not sitting in a pilot seat. My exact situation does involve a military flight where the mission is specifically to train multiple copilots. I’m not concerned about my own military rules for logging time in this forum thread, just the FAA rules regarding logging the time. Thanks all.

You didn't say what branch of service you were in, but in the hard cover logbooks used by the USN and USMC, there are 4 columns under the header "PILOT TIME".  Those 4 columns are listed below, along with a brief explanation of what I remember we would log in each category:

TOTAL PILOT TIME - (The sum of the other three columns - note that it isn't called total "FLIGHT" time...)
FIRST PILOT - (Time logged as manipulator of the controls.)
CO-PILOT - (Time spent not as manipulator of the controls, but may be acting as navigator, talking on radios, instructor letting student fly, etc.)
A/C COMMANDER - (Self explanatory, similar to PIC in that you can act as A/C commander even while not manipulating the controls, so time in this column may or may not match either of the other two columns.)

These categories are roughly equivalent to FAA definitions of PIC and SIC time, but they obviously aren't an exact match.  Clearly though, total time is ONLY intended to include pilot time.
There is a separate column, NOT under the "pilot time" header, labeled SPECIAL CREW TIME.  This column was used to log time for other crew members such as non-pilot rated officers, (Naval Flight Officers) or enlisted crewmembers such as flight engineers and loadmasters.  It could also be used by rated pilots in case they wanted to log total time in the airframe while not occupying a pilot seat or performing pilot duties.  (such as dead heading in a back seat)  Many of my pilot friends purposefully avoided logging crew time to keep their logbooks "pure".
Since you cannot log pilot time when not occupying a pilot seat, and since there is no category for total flight time, the question really becomes, what is your operations officer wanting to do with "TOTAL" time?
Because if you stick with the way the logbook is laid out, then total time will only include pilot time.  If you want to track special crew time for any reason, then log it.  It probably won't count for pilot upgrade purposes, (depending on policy) but it might be good enough for your local Lockheed Martin representative to present a 1000 hour pin.
However, if you want to somehow differentiate between reading/sleeping in the cargo bay on long flights vs sitting in the flight engineer seat observing and actively participating in quality training, (which has tremendous value, and I think is really the root of your question...) then establish a policy to only log crew time when directly observing the two pilot seat occupants.
Alternately, you could choose to log all non-pilot "flight time" in the airframe time as crew time, but create a separate training spreadsheet to track non-pilot seat, active monitoring of the PIC and SIC, with the A/C Commander signing off the training log after each flight.

@michaelhall Thanks so much, sir. Your post set me straight with understanding how the FAA regulates the logging of time in this situation. I truly appreciate everyone’s help. This is in regards to a Coast Guard unit, btw. Our Ops will, just as you said, count this ‘special’ time towards minimums required for right seat copilot to left-seat upgrade.  Again, thanks all. Question resolved.

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sort: votes


text:
What is altitude above surface / ASFC?
ASFC is an acronym frequently seen in official European publications, e.g. used by the French AIS:

Also in some ICAO ones, e.g in Annex 4 (Aeronautical Charts) to the Chicago Convention:

The airspace is shown only from the surface to the higher of the
  following two levels: 1500 m/above mean sea level (AMSL) — 600 m/above
  surface (ASFC).

What is ASFC compared to AGL? Is ASFC also used over water?

"Mean Sea Level" is an average altimeter setting based entirely on barometric pressure. If you took an airplane and flew it down to 0 over a calm open ocean, you may either be swimming or flying, so AMSL and ASFC are not technically the same over sea. Think about tides... I can't think why AGL and ASFC would be different though, other than AGL on maps usually the highest point in a specific area.

I've never heard of ASFC before, and it doesn't appear in any FAA materials. In fact, the only hits for it in Google are in French, where it's used as a synonym for AGL. Can you clarify where/why you've seen that ASFC and AGL are considered to be different things? Without that, it's hard to see what the actual question is here.

Like Pondlife says, it would be nice to see the source of your diagram. I have a database of all of the Glossary terms from FAA publications, the FARs, and miscellaneous sources. Of the 5379 terms, none of them are ASFC.

Do you know what the letters are an abbreviation for in French? I’m wondering if this is referencing the geode? http://www.esri.com/news/arcuser/0703/geoid1of3.html

@JScarry, French documents use the English wording only. It don't think the geoid is involved, the geoid doesn't perfectly match ground level, and for seas, it's below MSL (below the ellipsoid).

As best I can tell, ASFC is only used in France and is synonymous with AGL.
First, it isn't in ICAO's list of abbreviations or the FAA's PCG, but it is in the French AIP in section GEN 2.2:

ASFC / Au dessus de la surface / Above surface

The AIP also lists AGL, but it doesn't explain the difference:

AGL   / Au dessus du niveau du sol / Above ground level

Second, the AIP uses both terms in a seemingly interchangeable way. For example, in ENR 1.2 (VFR rules). Compare this from 4.7...

Come over the aerodrome, perpendicular to the runway at 2000 ft AGL,
  lights on, then join the circuit for landing

...with this from 4.10:

Fly over the runway at 2000 ft ASFC, perpendicular to the runway
  heading with landing lights switched on,

I can't see any reason why there would be any difference in meaning in those two phrases, and the airports in those sections are not particularly near the sea.
Finally, if you consider Wiktionary a valid reference, the definitions for ASFC and AGL are almost the same and they're both stated as the 'opposite' of MSL:

AFSC (Aviation) Désigne une hauteur au-dessus de la surface du sol ou
  de l'eau par opposition à une hauteur AMSL, dont la référence est le
  niveau moyen de la mer.
AGL (Militaire) Above ground level : « Au-dessus du niveau du sol »,
  altitude relative au sol, par opposition à l’altitude absolue, mesurée
  par rapport au niveau de la mer.

Although this is a guess, I suspect that ASFC is simply an example of French phraseology that has now been superseded by English but hasn't been updated everywhere. I can't find ASFC used anywhere in a non-French context.

Please note that (as at January 2018) AFSC is still used on French 1:500 000 charts, most especially where it relates to the boundaries of military low-level flying corridors (AZBAs) and TMAs. It seems to be very much a French thing, but then, when in France ......

@StephenParker You mean the "OACI" chart set? (also here)

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text:
Is it possible to tilt a propeller for attitude control, similarly to jet thrust vectoring?
Are there examples of propellers tilted in a controlled way to help in attitude control for pitch and yaw, adding a torque component?

Tilted propeller assumption
This would be similar to thrust vectoring in jets.

Vectored thrust principle
Alternatively could a swashplate similar to the one used in helicopter main rotor be used to generate asymmetric thrust?
(I think this excludes convertiplanes where the propellers are converted into rotors to take-off, land or move vertically.)

I don't know of any full scale planes that do this but it's quite common in RC.

Here's one I built several years ago: https://www.youtube.com/watch?v=TDFUr_UDVTE

Note that gyroscopic precession would cause the sum total of torque forces to be different than the simple addition of the new thrust angle.

@JonathanWalters: A point I had forgotten. Would the gyroscopic torque stop when the propeller stops swiveling?

The force moment of the gyroscopic precession is a result of the change in rotation plane of the propeller, so as the swivel stops so does the force moment, so yes, sort of. In actual practice the result of the gyroscopic precession usually results in some oscillations, depending on the level of dampening, which extend beyond the initial change in rotational plane.

Yes, on airships.
The Zeppelin NT can rotate its two forward and one rear propellers to create lift instead of thrust, a capability of particular value when the airship descends in an atmosphere with a strong temperature gradient. From this page about the Zeppelin NT:

The two forward propellors swivel to 120° and the aft one to 90°, and
  a fixed aft lateral propeller give a high level of manoeuverability,
  and enable smooth, quiet flight and impressive fuel economy. Some of
  the unique flight characteristics come from the swivelling propellers,
  like vertical take offs, precise landings, hovering over exact
  locations, turns on its axis and flying backwards.

Tail propellers of the Zeppelin NT Airship. While the sideways mounted propeller is fixed, the rear propeller can swivel through 90° (picture source)

Front view of the Zeppelin NT with the forward propellers tilted up (picture source)
Another example would be the V-22 Osprey, obviously, but you might prefer to call those whirly things rotors, not propellers.

Correct, V-22 Osprey is not in my scope, Zeppelin NT is. I've added a note in the question. Thanks.

While in case most VTOL aircraft, the rotors are tilted, one unique case is the Curtis Wright X-19, which had what the company called tilt-propellers. Curtis Wright at that point was a propeller manufacturer and its designers made the aircraft around the propeller.

Image from warbirdinformationexchange.org
This in turn, was developed from the X-100

Image from airandspace.si.edu
The designers used a property of the propeller called the radial lift force:

... as a propeller is inclined towards the vertical from the horizontal, the resultant of the propeller's thrust and the pressure of the relative wind acting on the rotating disk is a force with an additional lift component in the vertical axis. ... short propellers with wide blades magnified the radial force effect by increasing the surface area of the propeller disk without the compressibility issues of longer rotor blades. This offered a potential advantage over other tilt-rotor models, such as the Bell XV-3, with longer and narrower blades that did not have sufficient surface area to take advantage of the phenomenon. The added lift generated from the radial force permitted an aircraft built with the specially constructed lifting propellers to have smaller wings, which decreased weight and high-speed drag. This resulted in the most aerodynamically efficient of all of the VTOL designs.

In hover, the roll control was provided by differential pitch of the propellers. The aircraft basically behaved like a high-disc loading helicopter and was statically unstable in hover and in pitch and roll at low speed. The aircraft crashed during its first transition flight in 1965, with the progra ending soon.

This method is also used in a number of small UAVs.

This is really helpful! I'm getting a 404 page not found from the called the radial lift force link though.

Yes, this is possible because many helicopters use this method to some of their control. There are also some aircraft like the V-22 Osprey which use some variation on this control method.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

@fooot I think this answer does address the question; it's not a critique, and it isn't requesting clarification.

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text:
Why did our flight avoid the Black Sea?
Just flew from Schipol Airport in the Netherlands to Changi in Singapore.  The inflight tracker looked like this:

Clearly there is a direct route over Odesa to Tehran.  So why would the flight path go "out of our way" and around the Black Sea?
EDIT:  Being from Australia, I didn't realise Odesa is actually in Ukraine, but I will leave the original question as is.

In 2014, flight MH17 from Amsterdam, en-route to Kuala Lumpur, was shot down by the Russian military during their clandestine operation in the east of Ukraine. Airlines are very cautious with active war zones since.
Odesa is in Ukraine, and is under frequent military attacks. The flight route from Odesa to Tehran would take the aircraft over Crimea, occupied Ukrainian territory and part of the war zone. Large part of the Black Sea is considered a risk area as well, due to Russian fleet operations.
Furthermore, due to sanctions, western airlines do not fly through Russian airspace.
The routing through Turkey is limited to a few routes, with limited capacity.
The results is the route your flight took.

Note that it avoided Afghanistan as well.

"due to sanctions, western airlines do not fly through Russian airspace" Not just western airlines. The whole thing with Russia seizing/confiscating leased planes means that a lot of non-western airlines (have to) avoid Russian airspace as well.

It might be worth expanding on the point about "limited routes through Turkey", as naïvely it would seem like a plane could fly across the Black Sea from (say) Constanța, Romania to Rize, Turkey and still avoid the war zone.

Also take in account that no overflights over Afghanistan are advisable due to the 'recent' regime change.

@MichaelSeifert War zones don’t have rigid boundaries, and one of the belligerents has a long history of shooting down civilian planes, so it’s safest to avoid the entire Black Sea.

@MichaelSeifert they could, and usually prefer to, take that route. The problem is that all flights from Europe to most of Asia want to take that route given the situation. This traffic is in addition to the significant traffic volume between US/Europe and the middle east (Dubai, Doha are major hubs) that funnels through the area as well. As a result there is more demand than capacity on that route. Given the choice of a delay or a slightly longer route, most airlines choose the slightly longer route.

Thanks, being from Australia, European geography is not one of my top subjects :P  I didn't even realise Odesa is in Ukraine.

Exactly @Jan, and Russia could well pull a Lukashenko and send fighter jets up to tell the jetliner to land on a pretense, and then "false alarm, y'all are free to go, here's a train ticket to Kazakhstan, the airplane is ours now" and it becomes spare parts.  Literally, Russia changed their internal law so they could seize foreign owned jetliners in country.

This is from flightradar24.com, just now. UTC 15:09, 14/06/2022

Everyone is avoiding the Black Sea, or at most, only crossing the southern 1/4.
The lone aircraft you see in the middle, circled in red is a USAF Global Hawk (so he doesn't count)

Hm, but isn't this view partly so extreme because of limitations with flightradar24's data? Surely there are at least a few Russian planes in the air somewhere around Volgograd, which just aren't captured by Western radar, nor do the Russians provide that information any longer.

@leftaroundabout  Zooming out in FR24 still shows a LOT of flights across Russia. Just a big empty hole over Ukraine and slightly East of that. https://www.flightradar24.com/multiview/49.81,29.74/4

@leftaroundabout, many Russian airports in the north-east part of the map are closed since the war began.

@leftaroundabout pretty much anything in the air there will be (para)military and indeed FR24 probably doesn't have that data (in fact it tends not not have that data for most military flights for most nations).

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text:
Is it always true that a laminar boundary layer will always have a smaller skin friction coefficient than a turbulent boundary layer?
For a particular Reynolds number, I was wondering whether a laminar boundary layer is always preferable in terms of skin friction. So I looked towards the equations:
Blasius, laminar:
$$
C_{fl} = \frac{0.664}{\sqrt{Re}}
$$
Prandtl, turbulent:
$$
C_{ft} = \frac{0.027}{Re^\frac{1}{7}}
$$
For High reynolds (>1e4) this is obviously the case, a laminar boundary layer will always have a lower friction coefficient ($C_{ft} \gt C_{fl}$). However at values lower than 1e3; $C_{ft} \lt C_{fl}$.
So my question is this: is there a situation where a turbulent boundary layer induced early will result in reduced surface friction than its laminar counterpart? If so, is this a practical thing to achieve, or is it near impossible?

Those approximations are meant to be used within the limits in which they are valid.

Take a look at the work of B.S. Stratford and how the turbulent boundary layer can be caused to separate with zero skin friction...

Turbulent boundary layers generate always more drag than laminar boundary layers, so a longer turbulent boundary layer will always give more drag than a shorter one. Or is the question actually if: "is there a situation where a turbulent boundary layer induced early will result in reduced total drag than its laminar counterpart"? In that case the answer is surprisingly yes


text:
Does a plane's service ceiling vary with season/temperature?
Aircraft use barometric altitude, e.g. FL300 is the barometric altitude of 30,000 ft MSL. In winter however, FL300 is actually lower than the absolute altitude of 30,000 ft because of a cold and therefore compressed atmosphere, and in summer FL300 should be higher than 30,000 ft because the heated up atmosphere stretches itself out.
Does this mean that in summer it would be more difficult for a plane to reach a certain barometric altitude (e.g. FL300) than in winter? Is it more likely for planes to fly at higher barometric altitudes / flight levels in winter because they don't have to fly up a way as long as in summer?

The published performance data is based on the International Standard Atmosphere, being 15°C and 1013 hPa at sea level and decreasing at a standard lapse rate with altitude until you're in the stratosphere where it's constant.
Stratospheric temperatures can be colder or warmer than the ISA temperature in the flight levels, and the time of year is a factor (also affecting the tropopause level - in the arctic in winter, the stratosphere may start in the teens).
So an airplane may have a service ceiling of FL410 at ISA but the actual flight level will be higher if temperatures are below ISA or lower if above ISA.
Performance charts will include charts for ISA deviations, for example ISA + 5, ISA + 10 etc. A hot summer day may be ISA + 15 in the stratosphere. The CRJ200, on an ISA +15 day could not get to its FL410 ceiling no matter how patient you were, letting it take its time, in mach mode. On the other hand on an ISA -10 day in winter, you might be able to coax it to FL420 it you're patient and careful, although flying right at ceiling is a dangerous place to be for the unwary.

There are a few misconceptions in this question that we need to address before it can actually be answered.
While there is a very slight seasonal variation in the average sea-level atmospheric pressure (and that pressure is actually slightly lower in the winter), weather phenomena such as high and low pressure cells, fronts, etc. have a much greater effect on atmospheric pressure than the time of year.
Now, on to the actual question. The "ease" with which an airplane climbs is related to the air pressure, not the absolute altitude. All else being equal, going from, say, FL200 to FL300 always takes the exact same amount of effort, regardless of the actual distance to the ground. (If the air is less dense than average, the airplane will burn a tiny bit more fuel going the longer vertical distance to reach the same flight level, so I guess you could say that that requires a little more effort, depending on how you define "effort".)


text:
What is the exact meaning of CAVOK and CAVU? How are they used by pilots and ATC?
Do CAVOK (ceiling and visibility okay) and CAVU (ceiling and visibility unlimited) have formal definitions worldwide?

CAVOK, source: formercaptain.ca
Are they used in ATC communications?

CAVOK is formally defined in ICAO Annex 3 ( APP3-2 2.2 Use of CAVOK ) . You can hear it at the airports where there is no ATIS , when ATC communicates the weather info.
CAVU is not formally defined in Annex 3 .

When the following conditions occur simultaneously at the time of observation:
 a) visibility, 10 km or more, and the lowest visibility is not reported;
 b) no cloud of operational significance;
 c) no weather of significance to aviation as given in 4.4.2.3 and 4.4.2.6;

information on visibility, runway visual range, present weather and cloud amount, cloud type and height of cloud base shall be replaced in all meteorological reports by the term “CAVOK”.

Could you quote the actual definition from the document you refer to?

So what may be the difference between CAVOK and CAVU?

This answers 1/2 of the question (while CAVU isn't defined by ICAO, it still has a meaning...)

CAVU means: "CLEAR OR SCATTERED CLOUDS AND VISIBILIY UNLIMITED (GREATER THAN TEN MILES)"
CAVOK means: "CLOUD AND VISIBILITY OK (METAR)."
This military document, has a little more background:

Overseas locations may use the contraction “CAVOK” (ceiling and
  visibility OK). CAVOK is used when there is no significant weather,
  the visibility is 10 km or greater, and the ceilings are greater than
  5,000 ft.

I have never heard them used in verbal ATC communication, and they are not listed in the Pilot/Controller Glossary so should not be used on the radio.

Actual Definition From Annex 3
When the following conditions occur simultaneously at the time of observation:
a) visibility, 10 km or more, and the lowest visibility is not reported;
Note 1.— In local routine and special reports, visibility refers to the value(s) to be reported in accordance with
4.2.4.2 and 4.2.4.3; in METAR and SPECI, visibility refers to the value(s) to be reported in accordance with 4.2.4.4.
Note 2.— The lowest visibility is reported in accordance with 4.2.4.4 a).
b) no cloud of operational significance;
c) no weather of significance to aviation as given in 4.4.2.3, 4.4.2.5 and 4.4.2.6; information on visibility, runway visual range, present weather and cloud amount, cloud type and height of cloud base shall
be replaced in all meteorological reports by the term “CAVOK”.


text:
Does the expression "stall speed" have a definition?
I read, not only on this site, that the stall speed of an airfoil doesn't exist, and I usually make the effort to stay away from this expression.
While an airfoil can stall at any airspeed, it's clear that the stall speed refers to the airspeed at which the stall occurs in an horizontal 1G flight.

Stall speeds for B747-200 (n=1). (Source).
This speed is the lowest one that can generate a vertical component of lift equal to the weight with the airfoil at its stall angle of attack.
I would compare this implicit assumption with the maximum airspeed of an aircraft: It's clear we are assuming at least an horizontal flight and a standard atmosphere, etc, else this speed can be anything too.
So what's wrong with this definition of the "stall speed"? Is the stall speed defined in the aerodynamics field?
Example of use: Stall Speed (Dr. Mustafa Cavcar).

Related

'Stall speed' is defined only with respect to the aircraft as it is the case where it has a physical meaning. We define stall as the point where the airfoil or wing stops producing the necessary lift due to flow separation (This is an extremely simple definition, which will suffice for now), i.e where the lift coefficient is maximum.
For aircraft, the stall speed is given as where the wing reaches the max. lift coefficient. Basically, we are using the stall speed because it is directly available to the pilot i.e. it is a derived quantity, based on the angle of attack, configuration etc. among others. On by itself, the value has no meaning, as you pointed it out yourself.
Now, coming to the airfoil, the same (or similar) definition cannot be used as there is no way of saying lift equals weight due to the nature of airfoil itself. Thus, the stall speed is a performance term, not an aerodynamic one.

Indeed, "the stall speed of an airfoil doesn't exist." But the stall speed of an airplane does exist, though it's dependent on the aircraft weight $W$, load factor $n_z$, configuration (e.g., flaps position) and the air density. This variability makes it disadvantageous (for pilots---even dangerous) to think of an airplane's stall in terms of a definite stall speed $v_{Stall}$, as if it were a constant characteristic of the type. What are (roughly) constant are the stall angle of attack and the stall lift coefficient $c_{L_{Stall}}$.
The equilibrium equation for the forces acting on the airplane In stall in its symmetry plane and perpendicular to its velocity is:
$$n_z\cdot W = \frac{1}{2}\cdot\rho\cdot v^2_{Stall}\cdot S_{ref}\cdot c_{L_{Stall}}$$
Doing the algebra, one can easily exspress the stall speed in terms of the remaining parameters ($\rho$ = density of air, $S_{ref}$ = reference area) The resulting expression will explain (i) that for an airplane, stall speed does exist; (ii) the nature of its dependences and (iii) that for an airfoil, to which force equilibrium does not apply, there is no "stall speed."

Hi DEL, I took the liberty of adding some MathJax to your answer. If you want to go back to the old version, go back into the edit history and revert to the original answer.

Hi, Peter. Thanks for your pains. I'm not yet familiar with writing equations here, and right now I have no time to learn. Hence I will not re-edit. Yet I do have issues with your editting: (1) Weight, W, is perfectly fine and I see no reason to change it to m*g; (2) I much prefer C_L_max to C_L_stall, as it helps understand what C_L_stall is; (3) The "remaining variables" must include W, n , and C_L_max, as well as rho, but not S_ref, which is a constant; (4) If one knows how to write equations here, a better idea would be to just write the "resulting expression," instead of writing about it.

The rest of the remaining variables is already explained in the text; I see no need to list them once again.

There is not a single stall speed, because stall depends on

Angle of Attack
Pitch rate
Mach number
Reynolds number
Load factor
Flap settings
Center of gravity position.

However, for regulations the simplification of a single stall speed is helpful; here it is the minimum speed at which the aircraft can be controlled with idle power. Note that the definition for stall speed in the US regulations for GA aircraft (Part23 subpart B §23.49) is quite vague, allowing for details like the flight mass and altitude being

in the condition existing in the test, in which V$_{SO}$ and V$_{S1}$ are
  being used;

At least the center of gravity needs to be in the position which results in the highest stall speed (normally most forward). The controllability condition can result in the stall speed not being at the highest lift coefficient, but at the elevator control limit. Note also that it is not determined in level flight, but with engines idle or providing no thrust.
Your linked example is rather theoretical - the lift curve slope of an airplane rarely looks so nice and clean like in the example. If you test a wing with a constant cross section in the wind tunnel, the resulting lift curve slope allows to define a maximum lift coefficient, and with this you can calculate a theoretical stall speed when adding wing loading and absolute size (for the Reynolds number). But the stall speed as measured in flight test will be different and most likely higher.

A stall speed can be predicted if the other variables that are factors for a stall at any given speed are known.
Obviously, the point of discussion with stall speed is to make sure everyone has awareness that it isn't a fixed number, and it is definitely not the authoritative variable regarding the point at which an airfoil will stall; that would be the angle of attack.
The reason we need to acknowledge the existence of an estimated stall speed during common, predictable flight conditions is because the airspeed is going to be one of the pilot's primary focus points during critical phases of flight where a stall presents a greater risk.
Attempting to utilize a speed is just more practical.  There are virtually unlimited stall speeds, so in the same way we refer to a number divided by zero as "undefined," stall speed is also numerically undefined.
Yes, it has a literary definition, but no, it doesn't have a numerical definition, just a reference point for various operational configurations in different aircraft.


text:
Safran 2S2 Engine Overhaul
Does overhaul of Safran 2S2 Engine reset all engine inspections? If engine TSO is 0 as of today, does it mean I can reset all 15/20/30/300/600/3000/3500 hours inspection to be 0?
Or I have to see Shop Visit report to see what work has been performed while overhaul?
My understanding is, we can't make all engine inspections 0, but, we have to see Shop Visit report to see what tasks they have performed on overhaul.
Thanks,

Under what aviation authority do you operate?

FAA, I would like to know how does it matter?


text:
How does aviation deal with satellite and Moon laser ranging, and other scientific lasers in the sky?
It seems there are many scientific laser ranging stations around the globe and some are mobile.

Scientists Bounce Laser Off Soviet Lunar Rover Lost On Moon Decades Ago. Source
Questions

How dangerous are those scientific laser beams for passengers and crew, and for the aircraft?
Is there a NOTAM or a temporary flight restriction associated with scientific laser activity?

Background
For satellite laser ranging:

In satellite laser ranging (SLR) a global network of observation
  stations measures the round trip time of flight of ultrashort pulses
  of light to satellites equipped with retroreflectors. This provides
  instantaneous range measurements of millimeter level precision which
  can be accumulated to provide accurate measurement of orbits and a
  host of important scientific data.

International Laser Ranging Service - Map of stations. Source.
and for the Moon:

Laser beams are used because they remain tightly focused for large
  distances. Nevertheless, there is enough dispersion of the beam that
  it is about 7 kilometers in diameter when it reaches the Moon and 20
  kilometers in diameter when it returns to Earth. Because of this very
  weak signal, observations are made for several hours at a time. By
  averaging the signal for this period, the distance to the Moon can be
  measured to an accuracy of about 3 centimeters.

For astrophysics purposes:

Adaptive optics systems require a wavefront reference source in order
  to correct atmospheric distortion of light. Sufficiently bright stars
  are not available in all parts of the sky [...], one can create an
  artificial guide star by shining a laser into the atmosphere.

See also

Thanks to Federico, a very nice timelapse video of laser activity at Mauna Kea (by Sean Goebel).

Just a comment because it's hearsay, but: I worked with a guy who had a degree in astrophysics.  He said that observatories that use artificial guide stars for their adaptive optics employ interns whose job it is to lay on a cot in the parking lot and hit a remote kill switch when an aircraft approaches.

Yes, I read that too, and Frederico good answer also mentions (sorry for the caps): "THE AREA WILL ALSO BE MONITORED BY OBSERVERS AND THE LASER BEAM WILL BE TERMINATED IF NONPARTICIPATING AIRCRAFT ARE DETECTED". Pretty boring job. Also read that small radars start to be coupled with the laser source. Thanks for the confirmation anyway.

If the issue is the beam itself, why not just have a sensor that watches in the area the beam is projected and automatically shut it off when movement is detected, no need to worry about a shivering intern losing focus for a moment. Or is it just that the telescopes are so expensive they don't want to spend money on extra equipment like that?

@mins You could use a high-resolution video camera with good nighttime capabilities, a good-quality motion-detection algorithm, and a computer to run it on and trigger the kill switch when it detects point motion in the camera's field of view. There would likely be some expense involved, but it seems like it might be slightly more effective than a freezing-cold intern having to stay constantly alert.

@mins Nope, and neither do I know the specifications of individual observatory telescopes for their high-quality imaging. I do know, however, that telescopes of that scale are quite expensive, so even a lesser system of similar quality for detection of motion near the path of the telescope's direction would likely be quite pricey as well. Hence my initial comment about cost. Though if a human can spot the planes moving, an equivalent camera probably wouldn't need anywhere near the same amount of sensitivity as an advanced telescope to accomplish the same task anyway.

@JAB: This may be feasible with a camera, however there are problems to solve. On one hand a significant field of view seems to be required, preventing a zoom to be used in the initial detection. On the other hand, without zooming, targets will not be detected in the visible spectrum. This is where human eyes are more effective with their peripheral vision: good sensitivity on a large field, good accuracy at the very center to improve information about the target detected on the periphery.

@mins That is a good point.

Is there a NOTAM or a temporary flight restriction associated with laser ranging activity?

Apparently yes:

The lasers on the telescopes are in the range of 15-40 watts. The FAA calls a no-fly zone over the area when a laser is in use, and two people have to stand around outside in the freezing temperatures and watch for airplanes. Each of them has a kill switch to turn off the laser in case an airplane comes near. [...]
To clarify, the FAA does not have a no-fly zone, but instead issues a “Notice to Airmen” or NOTAM about the laser operations.

I also worked on an Adaptive Optics project and it was given for granted that NOTAMs would be issued by the people in charge of the experiments.
Example of NOTAM:

FDC 0/7651 ZDC .. SPECIAL NOTICE .. LASER OPERATIONS EFFECTIVE
1005010001 UTC UNTIL 1112312359 UTC. SCIENTIFIC AND RESEARCH LASER
OPERATIONS WILL BE CONDUCTED AT THE NASA GODDARD SPACE CENTER IN
GREENBELT, MARYLAND, LOCATED AT 390113N/764939W. THE SYSTEM IS
INTERMITTENT, WITH POSSIBLE OPERATIONS HAPPENING 24 HOURS A DAY, 7
DAYS A WEEK. THE LASER BEAM MAY BE INJURIOUS TO PILOTS/AIRCREWS AND
PASSENGERS EYES FOR A DISTANCE OF 98,000 FEET ABOVE GROUND LEVEL.
HOWEVER, THIS SYSTEM USES A LASER HAZARD REDUCTION SYSTEM RADAR THAT
IS SLAVED TO THE TELESCOPE MOUNT TO ENSURE THE LASER IS DEACTIVATED IN
THE EVENT AN AIRCRAFT APPROACHES. THE AREA WILL ALSO BE MONITORED BY
OBSERVERS AND THE LASER BEAM WILL BE TERMINATED IF NONPARTICIPATING
AIRCRAFT ARE DETECTED. LASER IRRADIANCE LEVELS WILL NOT EXCEED THE
MAXIMUM PERMISSIBLE EXPOSURE LEVELS WITHIN THE LASER FREE, CRITICAL
FLIGHT, AND SENSITIVE FLIGHT ZONES. OTHER VISUAL EFFECTS, E.G.,
FLASHBLINDNESS, AFTER IMAGE, GLARE, AND DISTRACTION MAY OCCUR AT
GREATER DISTANCES. THE POTOMAC TRACON /PCT/ PHONE NUMBER 540-349-7541
IS THE FAA COORDINATION FACILITY.

How dangerous is a laser beam for passengers and crew, and for the aircraft?

For the aircraft (meaning the hull) it is no danger, 40 watts of unfocused power (astronomers do not need laser beams in the sky as focused as the beams in an optics lab) do not affect the hull.
For the eyes of pilots and passengers it can instead be extremely dangerous, see Why is laser illumation of a cockpit an emergency?

"At one telescope, planes get close enough to the beam to cause a shutoff once every year or twin". Well! -- I'm adding a link to the Mauna Kea timelapse which is a pure beauty.

@mins note that that is not used for ranging, but for Adaptive optics systems: https://en.wikipedia.org/wiki/Adaptive_optics

I WISH THEY WOULD STOP WRITING IN ALL CAPS SO THAT WE COULD ACTUALLY READ THE NOTAMS. :)

Teletype machines were caps only. Aviation habits change quite slowly.

So the area around these laser ranging sites basically becomes a "controlled firing zone"; not marked as a potential hazard area on sectionals (though the observatories themselves, typically placed on mountains in the middle of nowhere, are common VFR landmarks) and non-participating aircraft don't have to avoid them, but it would be nice if you did.


text:
What is the exact responsibility of the Landing Signal Officer on an aircraft carrier?
From Youtube:

There are three LSOs apparently working at the same time, providing indications to the pilots:

What is the exact role of the LSO?
Why three? All of them holding a wave off switch.
What is the reason to have pilots guiding pilots rather than a computer?

Related: What is the procedure for landing on an aircraft carrier?

An Landing Signal Officer (LSO) is a naval aviator with additional specialized training to better facilitate recovery operations on the ship.  LSOs provide guidance for aircraft making approaches to the carrier.  They monitor the approach and remain in contact with the pilot during the approach by radio.
Carrier approaches or 'passes at the boat', while analogous in technique to an approach to land at a terrestrial airport, require much more precision and have far less margin for error due to the landing area's small size (75 x 600ft). And the requirement that the jet must impact the deck on speed and on angle of attack within a small area to snag an arrester wire and trap successfully makes this even more difficult.
The Navy has adopted this policy of the landing signal officer as well trained LSOs can quickly dissect problems with the approach and alert the pilot to correct prior to the pilot even becoming aware that there is a problem developing.  Particularly insidious errors like settling in the groove or settling at the ramp leading to a dangerous rampstrike or liquid landing can be spotted by the LSO before the pilot even realizes its happening.
In addition to forcing each pilot to perform at their best, the LSO grades each pass.  Multiple LSOs observe the approach and comment on the elements of the approach e.g. Long in the Groove, a little fast, low at the ramp, etc.  These are used to assign the following grades to a pass.

OK UNDERLINE - perfect pass, only awarded in extreme circumstances or
emergencies. Worth 5 points.
OK - excellent approach. On lineup, speed and glidepath with minimal
deviations, each corrected for quickly.  Hook snagged the 3-wire.
This is the best grade assigned for normal operations. Worth 4 points.
(OK) - A fair pass, exaggerated deviations with delayed corrections,
but approach was safe and conducted properly.  Worth 3 points
OK-BOLTER - an an OK or (OK) pass where the tailhook did not snag a
cross deck pendant.  Worth 2.5 points.
NO GRADE - Approach was safe but had gross deviations and
unacceptable corrections. Worth 2 points
WAVEOFF - Approach terminated, either by the LSO or the pilot for a
variety of unsafe reasons e.g. foul deck, PIO, etc.  worth 1 point.
CUT PASS - Unsafe approach attempted by the pilot, generally after
refusing to comply with a waveoff instruction.  If you survive it,
you can probably expect to have your wings clipped in a Field Naval
Aviator Evaluation Board.  Worth 0 points.

Grades do matter.  A single bad landing or a bolter will get you a razzing by your squadron mates; habitual bad landings and low grades will get you a FNAEB and most likely reassignment to flying helos or to shore based aircraft.
At least four LSOs will be on hand to observe and grade landings.  This 'Wave Team' will consist of a Controlling LSO, who monitors glideslope and speed, ultimately assigning the grade, a Backup LSO, generally more senior than the Controlling LSO who monitors lineup and assists with grading, and a Deck Status LSO, who monitors the condition of the deck in order to provide a waveoff alert to both the Controlling and Backup LSOs.  A fourth LSO, the Airwing LSO 'Paddles CAG' will be on hand to observe and supervise the whole operation and mentoring junior level LSOs.  There are two Paddles CAGs on every cruise and at least one is on the platform for every recovery.
In the photo above you have a group of LSOs observing approaches during a recovery.  The wired wands that two of them hold are controllers for the waveoff and cut lights on the IFLOLS optical landing aid system to the left of the landing area.  These give visual cues of mandatory instructions from the LSO for a waveoff or for the pilot to cut their engines on the approach.  Each LSO is noting where the aircraft was during the approach vs where it should have been to get a consensus as to what grade and comments to assign to an approach.

That's a lot of interesting elements, thanks. Would this procedure still apply during actual war operations?

The procedure remains the same aux pax aux bellum.

Excellent answer!  Just to clarify and address the original question, only the primary and backup LSO hold a "pickle" or waveoff button, not three as was questioned.  Also, there is always a bookwriter on the team.

Interesting that the pilot gets penalized for a waveoff, particularly if there's something fouling the deck, which is totally out of his control.

Foul deck waveoffs don't count against your landing grade.


text:
How is wind direction reported (blowing from or blowing to)?
When the tower controller, the ATIS or a METAR says Wind 270, 15 knots does it mean the wind is blowing in the westward or in the eastward direction?
Is there a single convention used to express the wind direction?

Wind direction in what? METAR?

@aeroalias: METAR, ATIS, or by the tower if that matters.

You know, I've been meaning to ask this for ages. Thanks for picking up my slack!!

@mins, I find it amazing that someone who asked such a seemingly basic question just 3½ years ago now is one of the top-players here on AvSE. This is just great.

@PerlDuck: Trivial questions are indeed/sadly among the most voted. QED.

I wouldn't call that sad. In hindsight these questions are trivial, but they weren't by the time they were asked. A good question is one that hasn't been asked before.

And when you take the time to research and put together an answer like this:  https://aviation.stackexchange.com/questions/12208/how-can-pilots-fly-inside-a-cloud/12217#12217, even the casual aviation "enthusiast" can quickly amass a lot of reputation points.

As far as METARs are considered, the wind direction gives the direction from which the wind is coming. From METAR definitions:

Wind Direction.
The direction, in tens of degrees, from which the wind is blowing with reference to true north.

So, Wind 270 shows that the wind is coming from west.
The reporting in ATIS and tower is the same, only difference being that the reference is magnetic north, while it is true north for METAR.

Great. What does "coming from" even mean... Is it pointing west, or is it pointing east? Raaahhh!

I think "Coming from the west" means "the air molecules are coming from the west".

Does this mean that a "Nor'easter" is coming from the North East?

@FreeMan Yes. At least on/near land it blows from northeast to southwest, hence the name. Though a Nor'easter is a cyclone-style swirling air mass, so the wind direction you feel depends on where in the swirl you are. But again, its name comes from the wind you feel on land, and that'll be a nor'easterly wind.

thanks, @Flambino. One of those things that seems obvious until you think about it a bit, then you start to wonder... :)

Wind numbers say where the wind is coming from. Wind 270 means that the wind is coming from the west, and blowing towards the east. If you point west (270), you will have the wind in your face as a headwind. If you point east (270 - 180 = 90), you will have a tailwind, or the wind at your back.
Another way to think about this is if you want to take off from runway 27, where you are pointing to 270, (pointing west), you want the wind coming from 270, so that you have a headwind.
It's easy to remember that when you are on runway 27 you are pointing to 270. Just remember that a headwind aligned with the runway will have the same number, and you will remember that the wind direction is where it is coming from.

Yet another way to think about it - an old fashioned weather vane will point into the wind, and it's an easy instrument to read

I always use the old saying;

The North wind doth blow, and we shall have snow...

Since this saying originated in England and in England it is colder to the North, we can conclude that a North wind blows from the North.
Incidentally, it is precisely the opposite with ocean currents. The Gulf Stream is a North-Easterly current, because if you are floating in it, it will take you to the North-East.

"opposite with ocean currents", interesting point indeed.

@mins it comes from the fact that for a boat, the direction the current is taking you is what you care about. For both a sailing boat and a plane, the direction the wind is coming from (and therefore passing over the sails/wings) is what matters. So we report both as the "most useful" version

As a pilot, you really want to know the wind direction on takeoff and landing. The reported direction will be where the wind is coming from, that is, opposite the direction the windsock is pointing. A good rule of thumb is that if the direction is reported in writing (METAR) then the bearing is relative to true north; if by voice (ATIS) then relative to magnetic north. This makes sense from the perspective that ATIS is related to a specific airfield, whose runway directions will be numbered according to the local magnetic variation. At an untowered field, pick the runway numbered closest to the ATIS-reported direction.

Carlo, it would be helpful if you provided clarification of your review.  How did I not answer the question?  If the asker has ever seen a windsock, then "coming from ... opposite the direction the windsock is pointing" seems like a pretty obvious mental picture.  Also, none of the other answers specifically articulated the "written vs. spoken rule", which is what I was searching for when I arrived here.

While not an absolute answer, or even trying to be, this answer provided some useful information and context that other answer did not.


text:
What does the actual air flow in a turbojet engine look like? Axial or helical?
Wikipedia describes the flow path as axial:

(Flow path, Wikipedia, author: Jeff Dahl)
This is the way the flow is depicted on many diagrams, however shouldn't the flow  be somehow helical:

This helical flow seems more accurate when looking at the details of the flow across rotors and stators:

So my question: How blades and vanes impact the air path?

What did you use to create the air flow path in your second image?

Wow good photo editing!!

Thank you, it's very well done and now I have a new toy to play with. :)

May I suggest you upload your image to Wikimedia and integrate it into the article? It would be rather nice addition.

@isanae The OPs image "looks nice", but unfortunately it is completely wrong, because, as another answer said, it ignores the effect of the stator vanes (which are not very obvious in the either illustration, but absolutely essential to how the compressor and turbine stages work).

@alephzero Fair enough. I have zero knowledge, I just found this on the HNQ and I was hoping we could improve the Wikipedia article. Many people discuss articles on SE without ever making corrections to them.

@mins This page http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node92.html and the following ones (click "next") are the best I can find on the web right now. The answer by chris is basically correct.

As @alephzero says, it's hard to see on those images but there is a static blade between each compressor stage called a stator blade. This removes the circumferential velocity and makes it axial again, ready for the next stage.

Ok, so there will be some rotation as per your image due to the stators not putting the flow back to its original circumferential position but I think it would be a lot milder. Rotating a blade pitch or two per stage. Bearing in mind you could have anything from around 28-30 blades to 50+ depending on the stage, the pitch isn't huge.

Interesting read here - NASA Compressor Page essentially says the flow is mostly axial and the job of the stators is to minimise global rotation around the engine. So they're saying there is some rotation as you describe but it is minimised as much as possible.

You should do a little bit of research on blade velocity triangles. You'll see that

Stators swirl the air
Blades de-swirl the air

So if you look at several rows of stator-blade pairs, the air flow will look like a stepped helix. This video has an alright explanation of the theory:

If you want to have a good look inside engines, I recommend the AgentJayZ channel on youtube, he does excellent "show and tell" of components in an engine though doesn't cover all theory very well, for example:

+1 that first video is exactly what is needed to answer this question.

Blades and vanes impact the flow such that the flow follows the paths of the cascades. The principle of a compressor is that the blades increase the angular momentum of the flow. Thus, the flow somehow resembles the screw in your drawing, but only for a few clockings. The vanes remove the angular momentum and concert it into inner energy of the fluid. Ideally the flow is in axial direction at the outlet of a vane. Blade and vane cascades are installed intermittendly, such that the flow goes axial - screw -axial - screw....
The turbine works the other way around. The vanes create angular momentum that is removed in the blade cascades.


text:
What type of variation in temperature does the wing undergo?
Generally what is temperature variation does the wing of the commercial aircraft like A380 undergo , flying at mach speed of 0.8 and how does it affect the aircraft other than icing , what type of literature can one go through to study the temperature effects on aircraft

Related: What is the typical temperature of an airliner's hull during flight?

The temperature rise that is caused by flying at high subsonic/transonic speeds is approximately 40 degrees Celsius. But this happens only at the stagnation point (near the leading edge). Ref: aerodynamics text books.
The A380 and other airliners experience -55 to -60 degrees Celsius during cruise at altitude. Ref: aerodynamics (properties of the atmosphere)
On the ground they may experience up to 60 degrees Celsius of air temperature and with the sun radiation the surface temperature can rise up to approx. 80-90 °C. Depending on color and geographic location on earth. Ref: hard experience :) and Environment control system Analysis.

90°C on the ground? That's only 10°C below water boiling point! Air temperature at 60°C at ground level, yes but this is very exceptional.

Ive given the extremes, but they are quite usual for some regions in the world.


text:
If indicated air speed does not change can the amount of lift change?
A friend and I were having a discussion. He says that if you are flying in a 50 knot headwind and the wind suddenly stops, the airplane will lose lift.
I say that this will not result in a loss of lift because, although the ground speed will change, the indicated airspeed does not change and no lift will be lost.
He then brought up microbursts and wind shear as an example of wind direction changing causing a change in lift, however I am not so sure. I think in those examples it is actual downdrafts pushing the plane down or the movement of air over the wing in a non-coordinated way, similar to how the air in a slip does not pass over the wing in a way that generates the most lift.
Does anyone have an answer? If an airplane is in a 50 knot headwind and the wind suddenly stops, will the airplane lose lift? What is the actual cause for loss of lift in a microburst or in wind shear?

You've got it the wrong way round. If the headwind stops the ground speed remains the same and the airspeed drops by 50 kt, so yes you loose lift (and I have had it happen on finals - rather scary).

The false assumption here is that airspeed won't change. But it will, it will drop by 50kt. If you kept the same engine power, it would get you back up to your original airspeed (assuming you were cruising), but not instantly.

Your friend is right.  Wind shear will greatly degrade or increase the lift depending on the change in direction and velocity of the shear.
If you are flying at 50 KIAS into a 50kt headwind, your ground speed will be zero knots.  If the wind shears to perpendicular to your flight path, your airspeed drops to zero knots and you fall out of the sky.  If it shears to 180deg in, says a microburst your airspeed is now -50kts or 50kts from the tail end almost certainly causing a departure from controlled flight.  In both cases at low altitudes the result is almost certainly fatal.
Conversely, if you were flying at 50KIAS and had a 50kt tailwind and the wind then sheared 180deg to a 50 kt headwind, you would abruptly jump to an airspeed of 100KIAS.  The airplane would rise very rapidly; you would feel the vertical acceleration in your seat.  This situation isn't as dangerous as shearing to a tailwind at low altitudes/speeds but it could be dangerous if one was cruising past Vno (yellow arc) where there is a risk of structural damage to the aircraft.
Your friend is right; microbursts are very dangerous.  The combination of wind shears as much as 90kts and severe downdrafts in the apex of the storm make it one of the most hazardous weather phenomena an aircraft can encounter.  They can even be catastrophic to large airliners at low speeds.  On August 2, 1985, a Lockheed L-1011 TriStar encountered a microburst while on final for KDFW and consequently crashed, killing 137 people.

As far as lift is concerned, it depends on the speed of the aircraft relative to the air surrounding it. The lift equation,
$L = \frac{1}{2} C_{L} \rho V^{2} A$
uses the true airspeed. This is the actual flight speed of an airplane relative to an air mass i.e. the atmosphere.
If an aircraft is flying into a 50-knot headwind, which suddenly stops for some reason, the lift will be lost (by an amount proportional to the square of the reduction in speed).
As far as micro-bursts and wind-shear are concerned, the effect will be dependent on the situation which the aircraft finds itself in. Windshear is a generic term for any rapidly changing air currents. As far as mirobursts are concerned, the aircraft first experiences a lift first and then loses it as it enters and exits the microburst.

Microburst, Image from propilotmag.com
From the NASA page on windshear and microbursts:

Windshear is a generic term referring to any rapidly changing wind currents. A type of weather phenomenon called "microbursts" can produce extremely strong windshear, posing great danger to aircraft. These are local, short-lived downdrafts that radiate outward as they rush toward the ground.
As a downdraft spreads down and outward from a cloud, it creates an increasing headwind over the wings of an oncoming aircraft. This headwind causes a sudden leap in airspeed, and the plane lifts. If the pilots are unaware that this speed increase is caused by windshear, they are likely to react by reducing engine power. However, as the plane passes through the shear, the wind quickly becomes a downdraft and then a tailwind. This reduces the speed of air over the wings, and the extra lift and speed vanish.

Approaching this from a physics point-of-view, it seems to me, that -
Note that the engines are propelling the aeroplane at say 200mph relative to the wind. The lift is based on this airspeed. However, a aircraft has a substantial inertia, which is relative the the ground (actually to the Universe, but the ground is a good substitute).
If the wind drops from 50mph to zero, then the ground-speed will remain at (200-50 mph), but the airspeed will fall from 200mph to 150mph. The lift will then drop appropriately.

To put a silver lining on this cloud, the drag will drop appropriately as well. :-)

The answer is yes.
Essentially the formula to obtain lift is the following:
$$
L=1/2\rho V^2C_L
$$
Where $$C_L$$ is the lift coefficient, which (in a simplified description) only depends on the configuration of the airplane (i.e. deploying flaps).
The density depends on the flying height and, to answer your question, the speed is the relative speed to the air.
That means, in the case of sudden change of external air speed it will change the relative speed (unless the airplane does not steers to compensate), so that the lift will change (decrease in this case).
A microburst brings increase and decrease of air speed reducing and increasing the lift. Wind shear will depend, but might increase of decrease speed depending on how affects.

As already mentioned, in short term it is your ground speed that remains constant. The first law of motion says

In an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a net force.

When the wind changes, it is accelerating, so it is not an inertial reference frame. Only the Earth reference frame is (after adding gravity).
When the airspeed changes, the aircraft will eventually resume the original airspeed due to the inherent stability or pilot or autopilot actions, due to the aerodynamic forces that act on it and that change with airspeed. This, however, takes time. So if the wind changes quickly, the ground speed remains and airspeed changes by the difference in wind speed.
Lift is proportional to angle of attack and square of airspeed. So if the angle of attack is reduced, so will be lift even if airspeed remains constant. And this can happen if the plane flies into a downdraft. It will initially continue straight (remember, first law of motion), but the air got own downward motion, so it hits the wings at lower angle of attack and the lift is reduced until the aircraft accelerates downward to the same vertical speed as the wind which restores the original angle of attack—or the pilot compensates with control input, but increasing angle of attack also increases drag and thus will cause the plane to slow down.

So when flying through a downburst, the airspeed will first increase due to headwind, which will increase the lift and make the plane climb. Then the wind will change to downdraft and this will pull the plane down. And last the wind will turn to tail wind, which will decrease the speed and make the plane continue descent. There is no updraft part, so the net effect is to put the plane below the original flight path.
But the most dangerous effect is stall. The lift increases with angle of attack only to a certain value, called critical angle of attack, when the wing will stall, which is accompanied by increase in drag and reduction of lift.
As the airspeed decays (due to downburst or windshear), the pilot will compensate by increasing the angle of attack to maintain the needed amount of lift. But if the airspeed drops too low, the angle of attack will exceed the critical angle, the wing will stall and lift will be lost. And due to the significant increase in drag, just adding engine power is generally not enough—the plane must pitch down and accelerate by trading some altitude for speed. Low above the ground the plane may not have enough altitude for this, which is why downbursts are so dangerous on approach and weather that generates them must be avoided.


text:
What is the power rating of the ground power unit for Bombardier Dash 8 Q400?
I am researching about ground power units for aircrafts and I need the power rating requirements (kW/kVA rating) of ground power units to supply electrical power to Bombardier Dash 8 Q400 and Embraer E145.
I know that aircraft need 115VAC and 28VDC, but I need the power rating of the GPU in kW or kVA like an Airbus aircraft needs a minimum of 90kVA GPU.

Specific information about requirements is difficult to find. However there is some information that is readily available concerning the Q400 and available ground power units.

The Q400 can start its engine with the onboard batteries, thus the necessary power requirement would not be nearly as big as with the A320 series.
Ground Power Units are usually supplied in 7,5, 15, 30, 60, 90, 120, 150 and 180kVA configurations
The Q400 is listed to have two TRUs which can output maximum of 300A 28VDC from a 115VAC (~73A) source. Source: Q400 Eletrical This would equal to a 90kVA Ground Power Unit, which according to Power Systems International provides 72A.
The Q400 has one 45kVA electrical generator per turbojet.
According to the Q400 Electrical document the Q400 has 24-volt NiCad batteries; Main and auxiliary, which both total to 80Ah.
The same document also states that the Q400 External Power Protection Unit does not monitor amperage, thus will not monitor total power. Some other system related with engine starting will most likely still monitor total power.
Aviation New Zealand provides the following details:

"Small business jets fitted with 115 V 3 phase systems may be supplied by lower
  power units, typically 7.5 kVA and 15 kVA. Aside from over-power limits, all other
  parameters are likely to be similar to the above. " above=minimum 45kVA (as according to the document is specified in
  ISO 6858:1982 (E): Aircraft Ground Support Electrical Supplies – General
  Requirements;)

Based on these details I think that the Q400 does not have specific external power requirements. Most likely a minimum of 7.5kVA is still recommended, as that is the lowest powered GPU I could find. However you might end up charging the batteries first, as the 7kW from the GPU might not be enough to start the engines. A 30kVA generator with 150% overload rating for 60 seconds is also most definitely enough since it can provide the exact 45kVA that is provided by the generators already in the engines of the Q400.
The best guestimate calculation I could come up of what the engines need to be started is:

A320 CFM56 provides about 30000hp max, while the PW100 provides about
  5000HP
      The CFM56 is started by a APS3200 APU, which provides 447hp, of which about 330hp can be used by the compressor equaling to about
  240kW.
      Based on the hp difference of about 15%, we will assume the PW100 requires
      only 15% of that power to start, which would be 36kW or ~40kVa.
      Of course not all of the 240kW provided by APS3200 would be required for a     CFM56 to start, so I'm pretty confident that a Q400 engine will start with a 30kVa GPU.

I think in the end it might come up to the question of using the batteries more, thus them needing more maintenance. NiCd usually rated to 1500-2000 recharge cycles, on Q400 and aircraft batteries it is most likely even lower.


text:
How does an airplane do a Kulbit manoeuvre?
Some time ago, I came across a GIF of a Russian military airplane doing a Kulbit manoeuvre (Probably for showoff). How, in terms of control surface movement, does an airplane to the Kulbit?

Source: Reddit

Thrust vectoring probably had a significant role to play.

@mins Readward salto

@mins It looks like the Kulbit so I will say that it is.

after seeing the video, I agree with Ron, without TV is not doable.

Upon seeing that clip, I can't help but recall this famous line: "Son, your ego is writing checks your body can't cash."

The Kulbit maneuver is a variation on the Cobra maneuver (Pugachev's Cobra) with a higher pitch rate and continuous positive pitch rate throughout the maneuver. It requires two things:

A docile pitch behavior of the airframe over the full angle of attack range of 360°. Docile means that the pitch moments stay low and do not change abruptly with the angle of attack.
Thrust vectoring nozzles to keep the pitch rate up.

An electronic flight system is helpful to make execution easier, but is not strictly required. Also helpful is an engine-intake-combination which does not stall the compressor if the outer flow is reversed briefly. As you can see, the control surfaces only play a minor role - they are only important initially to start the pitch-up motion. This is done by negative elevator and/or positive canard deflection.


text:
Can you have a Hybrid Turboprop-Jet engine?
To start things off I am writing this to gain information for vehicles in a story, so I am more so looking for theoretical possibilities, unless there are real world examples.
The background and my current understanding: Turboprops have great low end thrust and are more efficient for VTOL and low airspeed flight with the downside of low altitude flight. Jet engines, even going so far as Scramjets, are for faster and higher altitude flights.
The goal is to create a typical science fiction dropship that can land on unprepared surfaces and fly into space where another thruster onboard will propel the craft in space (unless there is a really diverse hybrid engine that could do that job).
So the key takeaways are the landing part and the high altitude. I know it is possible to take something like the V-22 Osprey and swap the Turboprops (Proprotors) for another higher thrust engine, but with the downside that you will burn and destroy any unprepared surface you land on.
The simple question: is it possible to have a Turboprop engine that essentially shifts to another form of jet engine for higher altitude and faster flights?
This is probably much more of a theoretical question than this particular site is used to so I am good with any helpful information.
Thank you

Do you want hard sci-fi (everything is actually physically possible with the laws of physics as we know it)? Is VTOL a hard requirement or a nice to have?  How much actually horizontal flying do you need to do? Is it just drop down from space to one site and then straight back up to space, or do you want to take off and fly around and land a few times?  Also, if you don't get answers here, Worldbuilding.SE may be a good resource for story questions.

Since it's sci-fi, have the props of the V-22 fold forward (away from the engine), then retract into the center of the spool to get them out of the way for "other" propulsion methods.

One idea I had was to fold the propellers back, then I wondered if the internal mechanics of the Turboprop could be "switched" to become a jet engine midflight.

@DanielK  somewhat hard, or at least theoretically possible with the use of future materials. The dropship would be used for many suborbital(atmospheric) flights with the ability to essentially become a space plane to make it back to their respected naval starship /space station.  So they would still need to be aerodynamically sound. So nothing like the dropship in The Expanse. A good comparison would be like the Pelican in Halo.

A key question to answer is how much oxygen and pressure are there in air the engine is supposed to fly. Are you assuming Earth atmosphere? And the second point is what is the initial speed (the speed of the mother ship)?

@mins yes,  I am assuming Earth atmosphere. As for the mother ship, at least at orbital speed. So up there between 8 to 10km/s+.

Are you assuming that you can refuel on the ground between sub-orbital flights?  Or is 100% of the fuel that you need for all flights going to have be carried when you initially drop?

Have you heard about the Rotary Rocket? It's a crazy hybrid of a helicopter and rocket that people made a fairly serious effort at building. Explanation and some footage of manned test flights: https://www.youtube.com/watch?v=OIuGfXp-Ok8

TLDR: Pretty much impossible with current technology as I understand the question.
I think you are focusing on the wrong part of the problem.  Turboprop vs turbofan is pocket change compared to the amount of rocket fuel required to get to orbit.  Let's put some numbers to it.
I'm assuming the following mission 1) Deorbit from low earth orbit, 2) landing (potentially parachute assisted), 3) a 10,000 km range for either one very long flight or a number of short ones, utilizing vertical takeoff 4) return to orbit.  Further assuming that all fuel for the entire mission is onboard at step 1.
We are going to start at the end, and work backwards.
First we need to know the mass of the vehicle. I'm assuming based on "typical science fiction dropship" that the actual vehicle weight might be similar to an F-35 fighter.  Let's take the empty weight of 13,290 kg (also similar to the v-22 osprey that you mentioned). Some sci-fi movies depict a "drop ship" as having a crew of dozens, so this might actually be on the low end, but let's start here. How much rocket fuel do we need to get that kind of a vehicle back into orbit?
Assuming the delta-v required is around 11,200 m/s, assuming a sporty specific impulse of 470s (the absolute best rocket listed in this table), and utilizing the rocket equation gives the fuel required as 150,000 kg.  For comparison, that's about the empty operating weight of a Boeing 777.  Now, that's assuming that you start from the ground and use the rocket the whole way.  Of course, if you use the jet engines to get started and launch from in-flight, you'd need a little bit less fuel, but it's not as much as you might think.  Elon Musk thinks its only about a 5% savings.  Let's just keep the 150,000 kg figure.
Now, we need to fly something with the weight of a B777 around a bit. For B777, a fuel load of 100,000 kg is enough to get around 10,000 km range,
So now our vehicle plus fuel is 250,000 kg. To do a vertical takeoff, you'd need a whopping 2.5 MN thrust, which in jet engine terms is around five GE9xs at max takeoff thrust (which by the way are about 10,000 kg each, so now our vehicle is way over our initially assumed weight, which means we need even more fuel!)
That's already a tremendous vehicle.  And we haven't even talked about how to get this thing to survive re-entry without burning up.  You'd need a massive heat shield, which would just add even more weight.  I would call this impossible with current technology.
So to modify it to make it work, you'd need to do one or more of the following

Massively cut back on the final mass.  i.e. make the final thing look more like an tiny Apollo command capsule than a "typical sci-fi drop ship".
Re-fuel on the ground before each flight and before returning to orbit.
Give up on VTOL
cut down on the flight range
give up on hard sci-fi, and invent a magic rocket engine with a tremendously better specific impulse than current technology.

Awesome answer, thank you for that. I figured it would be virtually impossible. And for the sake of the story I have contemplated something like a Sky Hook from the orbiting ship. That seems a little slow for the future setting. To give some more perspective since I think you settled the impossibility: the setting is about 600 years in the future where interplanetary travel is a few days only with torch drives similar to the Expanses Epstein drive.  The idea I am toying around with is to get the craft to a high enough altitude to turn on the fusion thrusters and get into space that way.

Well if you're going to go the route of "fusion thrusters" and "epstein drives", I don't think I'd worry too much about whether a turbofan is a bit less efficient than a turboprop at low speeds.  That's like saying you've got a Ferrari that can go 150 mph, but only if you tow it out to the highway with some draft horses first, and you are wondering what kind of horse would be the most efficient for towing. Just say that you have something like a jet engine but it runs off energy from your "fusion reactor" instead of jet fuel.

or to put it a different way, the reason we care about the efficiency difference between a turboprop and a turbojet is because jet fuel is expensive and heavy.  The specific energy of jet fuel is about 40 MJ/kg.  The specific energy of hydrogen fusion is 640,000,000 MJ/kg, so about 16 million times more energy per kg than jet fuel.  You could fly a B777 from New York to Tokyo on about 6 grams of hydrogen if you have fusion available.  So once you have fusion, turboprop vs turbojet is asking about whether you need 6 grams of hydrogen or 8 grams.  Not a big deal.

That's fair. I guess my question was more of a mechanical aspect on whether an engine that runs a propeller could also run a jet engine and compress the air and use that as the propellant in atmospheric flight.  The next phase I need to figure out is at what elevation a fusion engine could be used without disastrous affects.

I guess I'm not sure what you mean by fusion engine.  If you have a fusion reactor, use it to generate electricity, and then use that electricity to turn an electric motor connected to a big fan or propeller.  A fusion powered quadcopter would work nicely at sealevel up to where the air gets pretty thin.

I've read that the best way to use a fusion reactor for an engine is to use the heat instead of the electricity. But when I say I fusion thruster, the one I am referring to is basically a pellet of fusion fuel (deuterium-helium 3 in this case) is shot out the back of the thruster where a series of inertial confinement lasers ignite the pellet and the thruster/craft is propelled by this explosion. A more advanced and micro version of the Orion project. The benefit to this system is it keeps all the heat outside the vessel which is important since we are talking about millions of degrees.

Any VTOL arrangement will put a lot of blast on the landing surface, so a practical, off the shelf solution would be the Pratt & Whitney J58 hybrid turbojet/ram jet to get you off the ground and into near space at around Mach 3, before switching to rockets.
Although it is possible to mount a prop on this type of engine and fold it, drag and turbulence (on the intake air) may render it impractical.  Better to blow a few things over with the turbojet (great movie special effects).  The turboprop will exhaust hot gasses as well, at a slightly lower velocity.
Another approach may be a VTOL version of the British Skylon ground to space aircraft.  Perhaps in your book it actually makes it into production.
But for vertical landings, it may work best to supplement the turboramjets with rockets.  Then you wind up with something more akin to what Elon Musk is working on with jets and a little more wing.
Although these "hybrids" are a popular sci-fi concept, in reality only a small portion of even orbital flight is beyond the atmosphere.  Once past the point of lifting or oxygen consumption for thrust, jets, wings, and propellers become so much dead weight.

The F-35 Lightning has an approach to VTOL that is similar to what you are requesting. There is a propeller (strictly speaking, a ducted fan) pointed straight up in the middle of the fuselage and has doors that open above and below it. It is powered by the jet engine and used during vertical takeoff and landing.


text:
What is the difference between "origin date" and "departure date"?
In different sources e.g. OAG's API:s (Application Programming Interface), there are both (scheduled) "departure date" and "origin/origination date". In general (so not specifically OAG's API) is there a technical difference between these two or are they synonyms? Correlated, does the same apply to "destination" and "arrival"?

The definitions are probably in Amadeus documentation, as OAG uses MIDT data (now Market Insights). That's more a travel.se question, or better a SO question.

Context and acronym definition are always nice to include.  For instance, does OAG = Aircraft On Ground, and API = Aviation Preflight Instruction?


text:
How much drag does a RAT deployment add?
How much drag does a RAT deployment add?
I was watching a simulator video of an A320 dual engine failure at low height during climb out and one step in the protocol was the pilot manually deploying the RAT.
Simulator Video
Since he doesn't manage to make it back to the runway from his 2000 ft engine flameout run I was wondering if not deploying the RAT would make any difference due to reduced drag?
Would the on-board batteries and windmilling engines be able to provide sufficient power for control surfaces?
Also, are there real life incidents of a 2000 ft (or lower) dual engine failure that made a safe return to the runway? Sully Sullenberger's A320 had reached approx. 3,000 ft, yet he was forced to ditch rather than return.
Does the energy management differ drastically from type to type? i.e. Among dual engined aircraft would some be easier to land from a 2000 ft dual engine failure than others? For typical climb profiles, weights etc.

Your second and third questions would probably be better standing on their own as additional questions, instead of tack-ons for this one. It would be easier for future generations to find them that way.

Was that video from a professional simulator or a PC simulator like FS X? Because as far as I can tell (e.g. http://www.oaviao.com/oaviao_novo/diretorio_aero/manuais_voo/airbus/A320_meta.pdf, page 3.3) RAT deploys automatically on A320. And if the video was not realistic in this, I would doubt realism of the FDM as well.

@JanHudec Looks Professional to me. I've added a link. At approx. 6:10 in the video you can see him do a "Manual ON for the Ram Air Turbine"

Yes, it does look professional. Note however, that the first time around he does not do it. Seems more like “do this just in case it didn't deploy already”.

@JanHudec From what you mentioned it seems the deployment is mandatory? How much is the drag penalty? Not much I assume?

@curious_cat, wait a minute, even very rough estimates take some time to calculate ;-)

Since he doesn't manage to make it back to the runway from his 2000 ft engine flameout run I was wondering if not deploying the RAT would make any difference due to reduced drag?

No, because it deploys automatically. Note that in the video in the first attempt the step is not done because there is less time. It is probably included in the checklist just in case the automation didn't do it's job.
Nevertheless, let's try to estimate the drag. All I could quickly find about the RAT is that the attached electric generator provides 5 kW, but that should be enough. We need to take into account the hydraulic pump and power losses, but the generator wouldn't be insignificant part of the load, so we could guess that the turbine does not have more than 50 kW of drag.
Now potential energy is $E_p = mhg$, differentiating by time we get $P = v_{vert}mg$ and solving for vertical speed $v_{vert} = \frac{P}{mg}$. Taking some “median” weight of 60 t (A320 has MZFW 62.5 t and MLW 66 t) and substituting we get:
$$v_{vert} = \frac{50\ \mathrm{kW}}{6\cdot10^4\ \mathrm{kg}\cdot 9.8\ \mathrm{ms}^{-2}} \approx 0.085\ \mathrm{ms}^{-1} \approx 17\ \mathrm{ft/min}$$
So the deployed RAT increases the rate of descent by less than 20 ft/min.
Now the video mentions the best glide speed would be 205 knots (I suppose it is for the 1+F configuration; for clean it sounds too low, though I don't have a reference for this). Since the L/D ratio is around 18, that would give the total descent rate about:
$$v_{vert} = \frac{205 \mathrm{knot}}{18} \approx 11 \mathrm{knot} \approx  1150 \mathrm{ft/min}$$
Except, well, 18 is clean L/D. The L/D at 1+F is a bit lower. I don't have reference for that either, but something like 10–12 sounds right, so:
$$v_{vert} = \frac{205 \mathrm{knot}}{12} \approx 17 \mathrm{knot} \approx  1730 \mathrm{ft/min}$$.
That is about 100 times more. So the RAT might increase drag by 1%. It was very rough estimate, so it may be 2% or 0.5%, but it definitely isn't that significant.

Would the on-board batteries and windmilling engines be able to provide sufficient power for control surfaces?

The training manual mentions that

If engine wind milling is sufficient, additional hydraulic power may be recovered.

without saying at what speeds that would be, but generally the speed for wind milling engine restart are significantly higher that the best glide speed. I'd say the necessary speed would be more than 250 knots.

Also, are there real life incidents of a 2000 ft (or lower) dual engine failure that made a safe return to the runway? Sully Sullenberger's A320 had reached approx. 3,000 ft, yet he was forced to ditch rather than return.

I can't find any other incidents involving dual engine failure on A320-family aircraft besides US Airways 1549. And I am not aware of any incidents on any other aircraft that would fit that criteria either.
It should however be noted, that flight 1549 likely had enough altitude to return to the runway—if the manoeuvre was executed perfectly. A small mistake and they'd crash into populated area since the airport is in the middle of a city. The Sullenberger's great airmanship was in that he chose the worse option that he was certain he can make over the better option where he was not.

Does the energy management differ drastically from type to type? i.e. Among dual engined aircraft would some be easier to land from a 2000 ft dual engine failure than others?

Principally it's the same for any type. But since turn radius increases drastically with speed, slower aircraft can turn faster and thus could return from lower altitude.
Here's a related question that discusses the manoeuvre; and it's dangers: Is it even remotely feasible to turnback a single engine aircraft with an engine failure?

Fantastic answer! Thanks!

@curious_cat, I added a link to related question which discusses the turning back in some more detail.

@JanHudec good call on the 10x multiplier for useful power extraction from flow - this is supported by Hoerner's Fluid Dynamic Drag and Fluid Dynamic Lift books (and another paper I can't find now)


text:
Is the emergency fluorescent floor lighting strip along the aisle an item on the MEL?
Was travelling on a Airbus A320 (night flight) and during the safety briefing noticed that the  fluorescent floor lighting strip referred to in the evacuation instructions was totally absent.  Both sides of the aisle.
I could see the marks on the floor carpet where it must have been earlier tacked on. Was removed and went unreplaced for whatever reason.
Is it OK for an aircraft to fly without this item? Seemed pretty essential to me for a safe evacuation in the dark.
Edit: I did try & take a photo but unfortunately didn't come out so well.

Did you get happen to get a picture?

@MikeFoxtrot: Yes! I did actually. I'll try & post. Not sure how nice it came out.

According to this document the answer would appear to be no if all lights were missing.

The 1/1 columns indicates the number installed and required for dispatch.

Thanks! I am going to email the regulator, just to see what they say. I will report back.

Those lights don't have to be mounted to the floor. They can also be seat mounted.
http://www.bruceind.com/index.php?option=com_k2&view=item&id=90:escape-path-lighting-systems&Itemid=189
http://www.astronics.com/_images/aircraft-safety/EPM%20033010.pdf

This is the AC that provides guidance on the requirements of the system. http://www.faa.gov/documentLibrary/media/Advisory_Circular/AC25.812-1A.pdf
The regulation is 14CFR 25.812

Interesting! Thanks. Maybe there were other lights that I did not see? Possible. Better safe then sorry though. I'll better report to the regulator & if there was a different system then no harm done. What do people think?